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(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells in series,the number of equivalent

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$0.27$

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(B) According to Faraday's second law of electrolysis,when the same quantity of electricity is passed through cells in series,the number of equivalents deposited is equal.$n_{Cu} 	imes (n	ext{-factor})_{Cu} = n_{Al} 	imes (n	ext{-factor})_{Al}$For $Cu^{2+} + 2e^- ightarrow Cu$,the $n	ext{-factor}$ is $2$.For $Al^{3+} + 3e^- ightarrow Al$,the $n	ext{-factor}$ is $3$.Given $n_{Cu} = 0.4$ moles.$0.4 	imes 2 = n_{Al} 	imes 3$$n_{Al} = \frac{0.8}{3} \approx 0.266$ moles.Rounding to two decimal places,we get $0.27$ moles.

Two electrolytic cells are connected in series containing $CuSO_4$ solution and molten $AlCl_3$. If in electrolysis $0.4$ moles of $Cu$ are deposited on the cathode of the first cell,the number of moles of $Al$ deposited on the cathode of the second cell is: (in $moles$)

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